Circuit-slide-2

# Circuit-slide-2 - 1 Power in Circuits Consider the input...

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Unformatted text preview: 1 Power in Circuits Consider the input impedance of a transmission line circuit, with an applied voltage v(t) inducing an input current i(t) . For sinusoidal excitation, we can write [ ] ( ) cos( ) ( ) cos( ) / 2, / 2 v t V t i t I t ϖ ϖ φ φ π π = = - - where φ is the phase difference between voltage and current. Note that φ = 0 only when the input impedance is real (purely resistive). 2 [ ] ( ) ( ) ( ) cos( ) cos( ) ( ) cos( ) cos(2 ) 2 p t v t i t V I t t V I p t t ϖ ϖ φ φ ϖ φ = = - = + - 0 0 c o s ( ) 2 V I φ 0 0 c o s ( 2 ) 2 V I t ϖ φ- The time-dependent input power is given by The power has two (Fourier) components: (A) an average value (B) an oscillatory component with frequency 2ω 3 The power flow changes periodically in time with an oscillation like (B) about the average value (A) . Note that only when φ = 0 we have cos( φ ) = 1, implying that for a resistive impedance the power is always positive (flowing from generator to load). When voltage and current are out of phase, the average value of the power has lower magnitude than the peak value of the oscillatory component. Therefore, during portions of the period of oscillation the power can be negative (flowing from load to generator). This means that when the power flow is positive, the reactive component of the input impedance stores energy , which is reflected back to the generator side when the power flow becomes negative. For an oscillatory excitation, we are interested in finding the behavior of the power during one full period, because from this we can easily obtain the average behavior in time. From the point of view of power consumption, we are also interested in knowing the power dissipated by the resistive component of the impedance. 4 0 0 0 ( ) cos( ) cos cos sin sin i t I t I t I t ϖ φ φ ϖ φ ϖ = - = + cos( ) cos cos sin sin A B A B A B- = + 0 0 0 0 ( ) c o s ( ) c o s ( ) c o s ( ) c o s ( ) s in ( ) s in ( ) P t V t I t V t I t ϖ ϖ φ ϖ ϖ φ = + (Re ) 2 0 0 0 0 Re ( ) cos( )cos ( ) sin( )sin(2 ) 2 Active al Power active Power V I P t V I t t φ ϖ φ ϖ...
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Circuit-slide-2 - 1 Power in Circuits Consider the input...

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